Question

What is the energy released in the fission reaction 1 0 n + 235 92 U → 131 53 I + 89 39 Y + 16 1 0 n 0 1 n + 92 235 U → 53 131 I + 39 89 Y + 16 0 1 n ? (The atomic mass of 131 I 131 I is 130.9061246 u and that of 89 Y 89 Y is 88.9058483 u)

Answer 1

Answer: The energy released in the given nuclear reaction is 94.99 MeV.

Step-by-step explanation:

For the given nuclear reaction:

`_(92)^(235)\textrm{U}+_0^1\textrm{n}\rightarrow _(53)^(131)\textrm{I}+_(39)^(89)\textrm{Y}+16_(0)^1\textrm{n}`

We are given:

Mass of
`_(92)^(235)\textrm{U}` = 235.043924 u

Mass of
`_(0)^(1)\textrm{n}` = 1.008665 u

Mass of
`_(53)^(131)\textrm{I}` = 130.9061246 u

Mass of
`_(39)^(89)\textrm{Y}` = 88.9058483 u u

To calculate the mass defect, we use the equation:

`\Delta m=\text{Mass of reactants}-\text{Mass of products}`

Putting values in above equation, we get:

`\Delta m=(m_U+m_(n))-(m_(I)+m_(Y)+16m_(n))\\\\\Delta m=(235.043924+1.008665)-(130.9061246+88.9058483+16(1.008665))=0.1019761u`

To calculate the energy released, we use the equation:

`E=\Delta mc^2\\E=(0.1019761u)* c^2`

`E=(0.1019761u)* (931.5MeV)` (Conversion factor:
`1u=931.5MeV/c^2` )

`E=94.99MeV`

Hence, the energy released in the given nuclear reaction is 94.99 MeV.